Problem: The lifespans of porcupines in a particular zoo are normally distributed. The average porcupine lives $19.4$ years; the standard deviation is $1.1$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a porcupine living longer than $16.1$ years.
$19.4$ $18.3$ $20.5$ $17.2$ $21.6$ $16.1$ $22.7$ $99.7\%$ $0.15\%$ $0.15\%$ We know the lifespans are normally distributed with an average lifespan of $19.4$ years. We know the standard deviation is $1.1$ years, so one standard deviation below the mean is $18.3$ years and one standard deviation above the mean is $20.5$ years. Two standard deviations below the mean is $17.2$ years and two standard deviations above the mean is $21.6$ years. Three standard deviations below the mean is $16.1$ years and three standard deviations above the mean is $22.7$ years. We are interested in the probability of a porcupine living longer than $16.1$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $99.7\%$ of the porcupines will have lifespans within 3 standard deviations of the average lifespan. The remaining $0.3\%$ of the porcupines will have lifespans that fall outside the shaded area. Because the normal distribution is symmetrical, half $({0.15\%})$ will live less than $16.1$ years and the other half $({0.15\%})$ will live longer than $22.7$ years. The probability of a particular porcupine living longer than $16.1$ years is ${99.7\%} + {0.15\%}$, or $99.85\%$.